Here's a parameterization of a plane: $\vec{v}(x, y) = (2x + y, 7y, -4y + x)$ What vectors are normal to the plane $\vec{v}$ ? Choose 2 answers: Choose 2 answers: (Choice A) A $(-9, 7, -14)$ (Choice B) B $(7, -9, -14)$ (Choice C) C $(9, -7, 14)$ (Choice D) D $(-7, 9, 14)$
Solution: The vector normal to the area element describes what's perpendicular to the surface, and the vector's magnitude represents the area of a tiny rectangle along the parameterization. We can take multiply the result by $-1$ to find another normal vector. $\text{normal to area element} = \dfrac{\partial \vec{v}}{\partial x} \times \dfrac{\partial \vec{v}}{\partial y}$ Let's calculate the cross product. $\begin{aligned} \dfrac{\partial \vec{v}}{\partial x} \times \dfrac{\partial \vec{v}}{\partial y} &= \det \begin{pmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \\ 2 & 0 & 1 \\ \\ 1 & 7 & -4 \end{pmatrix} \\ \\ &= -7 \hat{\imath} + 9 \hat{\jmath} + 14 \hat{k} \end{aligned}$ We want two normal vectors. Because the negative of a normal vector is also normal to the surface, we can take the negative of what we just calculated as a second normal vector to the plane $\vec{v}$. Therefore, two vectors normal to $\vec{v}$ are $(-7, 9, 14)$ and $(7, -9, -14)$.